MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable CalculusFall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � � 1 218.02 Lecture 14. – Thu, Oct 11, 2007 Handouts: PS5 solutions, PS6, practice exams 2A and 2B. Non-independent variables. Often we have to deal with non-independent variables, e.g. f(P, V, T ) where P V = nRT . Question: if g(x, y, z) = c then can think of z = z(x, y). What are ∂z/∂x, ∂z/∂y? Example: x 2 + yz + z3= 8 at (2, 3, 1). Take differential: 2x dx + z dy + (y + 3z 2) dz = 0, i.e. 4 61 6dy. 4 dx + dy + 6 dz = 0 (constraint g = c), or dz = So ∂z/∂x = −4/6 = −2/3 and dx − − −1/6 (taking the coefficients of dx and dy). Or equivalently: if y is held constant then we substitute dy = 0 to get dz = −4/6 dx, s o ∂z/∂x = −4/6 = −2/3. In general: g(x, y, z) = c gx dx + gy dy + gz dz = 0. If y held fixed, get gx dx + gz dz = 0, i.e. ⇒dz = −gx/gz dx, and ∂z/∂x = −gx/gz. Warning: notation can be dangerous! For example: f(x, y) = x + y, ∂f /∂x = 1. Change of variables x = u, y = u + v then f = 2u + v, ∂f /∂u = 2. x = u but ∂f /∂x =� ∂f/∂u !! This is because ∂f/∂x means change x keeping y fixed, while ∂f/∂u means change u keeping v fixed, i.e. change x keeping y − x fixed. ∂f When there’s ambiguity, we must precise what is held fixed: = deriv. / x with y held ∂x � � y ∂f fixed, = deriv. / u with v held fixed. ∂u v ∂f ∂f ∂f We now have = = . ∂u ∂x �∂x v v y In above e xample, we computed (∂z/∂x)y. When there is no risk of confusion we keep the old notation, by default ∂/∂x means we keep y fixed. ∂z/∂y = Example: area of a triangle with 2 sides a and b making an angle θ is A = 1 2ab sin θ. Suppose it’s a right triangle with b the hypothenuse, then constraint a = b cos θ. 3 ways in which rate of change of A w.r.t. θ makes sense: ∂A 1) view A = A(a, b, θ) independent variables, usual = Aθ (with a and b held fixed). This ∂θ answers the ques tion: a and b fixed, θ changes, triangle stops being a right triangle, what happens b to A? 2) constraint � a� = b cos θ, keep a fixed, change θ, while b does what it must to satisfy the constraint: ∂A . ∂θ a 3) constraint � a� = b cos θ, keep b fixed, change θ, while a does what it must to satisfy the constraint: ∂A . ∂θ How to compute e.g. (∂A/∂θ)a? [treat A as function of a and θ, while b = b(a, θ).] 11 a 2 tan θ, (∂A ∂θ )a = 1 2a 2 sec2 θ. (Easiest 0) Substitution: a = b cos θ so b = a sec θ, A = ab sin θ = 2 2here, but it’s not always possible to solve for b) 1) Total differentials: da = 0 (a fixed), dA = Aθdθ + Aada + Abdb =1 2ab cos θ dθ + b sin θ da + 1 2a sin θ db, and constraint ⇒ da = cos θ db − b sin θ dθ. Plugging in da = 0, we get db = b tan θ dθ 1� � � � � � � � � � � � � � � � � � � � � � 2 and then 1 1 ∂A 1 1 1 dA = ( ab cos θ + a sin θ b tan θ)dθ, = ab cos θ + a sin θ b tan θ = ab sec θ. 2 2 ∂θ 2 2 2a 2) Chain rule: (∂A/∂θ)a = Aθ(∂θ/∂θ)a + Aa(∂a/∂θ)a + Ab(∂b/∂θ)b = Aθ + Ab(∂b/∂θ)a. We find (∂b/∂θ)a by using the constraint equation. [Ran out of time here]. Implicit differentiation of constraint a = b cos θ: we have 0 = (∂a/∂θ)a = (∂b/∂θ)a cos θ − b sin θ, so (∂b/∂θ)a = b tan θ, and hence � � ∂A 1 1 1 = ab cos θ + a sin θ b tan θ = ab sec θ. ∂θ 2 2 2a The two systematic methods essentially involve calculating the same quantities, even though things are written differently. 18.02 Lecture 15. – Fri, Oct 12, 2007 Review topics. – Functions of several variables, contour plots. – Partial derivatives, gradient; approximation formulas, tangent planes, directional derivatives. Note: partial differential equations (= equations involving partial derivatives of an unknown function) are very important in physics. E.g., heat equation: ∂f/∂t = k(∂2f/∂x2 + ∂2f/∂y2 + ∂2f/∂z2) describes evolution of temperature over time. – Min/max problems: critical points, 2nd derivative test, checking boundary. (least squares won’t be on the exam) – Differentials, chain rule, change of variables. – Non-independent variables: Lagrange multipliers, and constrained partial derivatives. Re-explanation of how to compute constrained partials: say f = f(x, y, z) where g(x, y, z) = c. To find (∂f/∂z)y: 1) using differentials: df = fx dx + fy dy + fz dz. We set dy = 0 since y held constant, and want to eliminate dx. For this we use the constraint: dg = gx dx + gy dy + gz dz = 0, so setting dy = 0 we get dx = −gz/gx dz. Plug into df: df = −fxgz/gx dz + gz dz, so (∂f/∂z)y = −fxgz/gx + gz. 2) using chain rule: ∂f = ∂f ∂x + ∂f ∂y + ∂f ∂z = fx ∂x + fz, while ∂z y ∂x ∂z y ∂y ∂z y ∂z ∂z y ∂z y 0 = ∂g = ∂g ∂x + ∂g ∂y + ∂g ∂z = gx ∂x + gz∂z y ∂x ∂z y ∂y ∂z y ∂z ∂z y ∂z y which gives (∂x/∂z)y and hence the
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